The sum of an infinite GP is x and the common ratio r is such that `|r| lt 1`. If the first term of the GPis2, then which one of the following is correct ?

Options

A. `-1 lt x lt 1`
B. `-oo lt x lt 1`
C. ` 1 ltxltoo`
D. None of these

Answer

To solve the problem, we need to find the range of values for \( x \) given that the first term of the infinite geometric progression (GP) is 2 and the common ratio \( r \) satisfies \( |r| < 1 \).

### Step-by-Step Solution:

1. **Identify the formula for the sum of an infinite GP**:
The sum \( S \) of an infinite GP can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio.

2. **Substitute the given values**:
In this case, the first term \( a = 2 \). Therefore, we can write:
\[ x = \frac{2}{1 - r} \]

3. **Consider the condition for \( r \)**:
Since \( |r| < 1 \), this implies two cases for \( r \):
- Case 1: \( r \) is positive (\( 0 < r < 1 \))
- Case 2: \( r \) is negative (\( -1 < r < 0 \))

4. **Analyze Case 1 (when \( r > 0 \))**:
- As \( r \) approaches 1, \( 1 - r \) approaches 0, making \( x \) approach infinity.
- When \( r = 0 \), \( x = \frac{2}{1 - 0} = 2 \).
- Therefore, in this case, \( x \) can take values from 2 to infinity:
\[ x \in [2, \infty) \]

5. **Analyze Case 2 (when \( r < 0 \))**:
- As \( r \) approaches -1, \( 1 - r \) approaches 2, making \( x \) approach 1.
- When \( r = 0 \), \( x = \frac{2}{1 - 0} = 2 \).
- Therefore, in this case, \( x \) can take values from 1 to 2:
\[ x \in [1, 2) \]

6. **Combine the results from both cases**:
- From Case 1, \( x \in [2, \infty) \)
- From Case 2, \( x \in [1, 2) \)
- Combining these intervals, we find:
\[ x \in [1, \infty) \]

### Conclusion:
Thus, the correct range for \( x \) is:
\[ x \in [1, \infty) \]